Answer

Verified

417.3k+ views

**Hint:**Since in the given question we need to determine the value of $y\left( { - \sqrt 3 } \right)$ so that means we need to find the value of $y(x)$ , in the first part we will be determine the value of $y(x)$ by integrating $\dfrac{{dy}}{{dx}}$ and it can seen that $y(x)$ involves the function $f(x)$ that means we need to simplify it get the value that can be substituted to get the required value of $y(x)$.

**Complete step-by-step answer:**

Given

$\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}\left( {f(x)} \right)} \right)$

Integrating both side

$\int {\dfrac{{dy}}{{dx}} = \int {\dfrac{1}{2}} } \dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}\left( {f(x)} \right)} \right)$

Integrating of any differential function $\dfrac{d}{{dx}}\left( {g(x)} \right) = g(x) + c$

So by using this

$y = \dfrac{1}{2}{\sin ^{ - 1}}(f(x)) + c$

Now we know that $f(x) = {\left( {\sin \left( {{{\tan }^{ - 1}}x} \right) + \sin \left( {{{\cot }^{ - 1}}x} \right)} \right)^2} - 1$

Now we made some modification in given above equation

Let assume ${\tan ^{ - 1}}x = \theta $ ( this assumption we made just to making easy to this equation)

And we know that

${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}$

From this we can written as ${\cot ^{ - 1}}x = \dfrac{\pi }{2} - {\tan ^{ - 1}}x$ and we assume ${\tan ^{ - 1}}x = \theta $

So by putting this thing in $f(x) = {\left( {\sin \left( {{{\tan }^{ - 1}}x} \right) + \sin \left( {{{\cot }^{ - 1}}x} \right)} \right)^2} - 1$

$f(x) = {\left( {\sin \theta + \sin \left( {\dfrac{\pi }{2} - \theta } \right)} \right)^2} - 1$

Using

$\because {\cot ^{ - 1}}x = \dfrac{\pi }{2} - {\tan ^{ - 1}}x = \dfrac{\pi }{2} - \theta $

Now, we know that $\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta $

So, $f(x) = {\left( {\sin \theta + \cos \theta } \right)^2} - 1$

Now, we know that ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$

${\left( {\sin \theta + \cos \theta } \right)^2} = {\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta $

Now $\because {\sin ^2}\theta + {\cos ^2}\theta = 1$

So from here we get

$\Rightarrow$$f(x) = 1 + 2\sin \theta \cos \theta - 1$

Simplifying the above, we get

$\Rightarrow$$f(x) = 2\sin \theta \cos \theta $

Using

$\because 2\sin \theta \cos \theta = \sin 2\theta $

we get

$\therefore f(x) = \sin 2\theta $

Now, put this value in

$y = \dfrac{1}{2}{\sin ^{ - 1}}(f(x)) + c$

we get

$\Rightarrow$$y(x) = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\sin 2\theta } \right) + c$

Now replace value of $\theta $ by ${\tan ^{ - 1}}x$

So,

$\Rightarrow$$y(x) = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\sin (2{{\tan }^{ - 1}}x)} \right) + c$

Now given that $y\left( {\sqrt 3 } \right) = \dfrac{\pi }{6}$

Hence,

$\Rightarrow$$\dfrac{\pi }{6} = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\sin (2{{\tan }^{ - 1}}\sqrt 3 )} \right) + c$

we get

$\because {\tan ^{ - 1}}\left( {\sqrt 3 } \right) = \dfrac{\pi }{3}$

By putting this value we get

$\Rightarrow$$\dfrac{\pi }{6} = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\sin \dfrac{{2\pi }}{3}} \right) + c$

Simplifying the above

$\Rightarrow$$\dfrac{\pi }{6} = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)} \right) + c$

We get,

$\Rightarrow$$\dfrac{\pi }{6} = \dfrac{1}{2} \times \dfrac{\pi }{3} + c$

From here

$c = 0$

Now our $y(x) = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\sin (2{{\tan }^{ - 1}}x)} \right)$

Now we have to find $y\left( { - \sqrt 3 } \right)$

So put $x = - \sqrt 3 $ in $y(x) = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\sin (2{{\tan }^{ - 1}}x)} \right)$

We get

$\Rightarrow$$y( - \sqrt 3 ) = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\sin \left( {(2{{\tan }^{ - 1}}( - \sqrt 3 )} \right)} \right)$

using

$\because {\tan ^{ - 1}}( - \sqrt 3 ) = - \dfrac{\pi }{3}$

By putting this we get

$\Rightarrow$ $y( - \sqrt 3 ) = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\sin \left( { - \dfrac{{2\pi }}{3}} \right)} \right)$

using

$\because \sin \left( { - \dfrac{{2\pi }}{3}} \right) = - \dfrac{{\sqrt 3 }}{2}$

By putting this we get

$\Rightarrow$$y( - \sqrt 3 ) = \dfrac{1}{2}\left( {{{\sin }^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)} \right)$

using

$\because {\sin ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right) = - \dfrac{\pi }{3}$

By putting this we get

$\Rightarrow$$y( - \sqrt 3 ) = \dfrac{1}{2}\left( { - \dfrac{\pi }{3}} \right)$

So it become

$\Rightarrow$$y( - \sqrt 3 ) = - \dfrac{\pi }{6}$

**So option A is the correct answer.**

**Note:**The best way to determine the specific value involving trigonometric function is to consider or to convert into the short term as in this the function $f(x)$ is simplified to short term that makes the solution extremely easy to solve otherwise it will be quite difficult to determine the result and if we need to determine the value of constant in that case we use the initial condition as in the above question.

Recently Updated Pages

Which of the following is correct regarding the Indian class 10 social science CBSE

Who was the first sultan of delhi to issue regular class 10 social science CBSE

The Nagarjuna Sagar project was constructed on the class 10 social science CBSE

Which one of the following countries is the largest class 10 social science CBSE

What is Biosphere class 10 social science CBSE

Read the following statement and choose the best possible class 10 social science CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Which are the Top 10 Largest Countries of the World?

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

Why is the Earth called a unique planet class 6 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

One cusec is equal to how many liters class 8 maths CBSE